\(\int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1456]

Optimal result

Integrand size = 29, antiderivative size = 104 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {3 a b \text {arctanh}(\cos (c+d x))}{d}-\frac {\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {3 a b \sec (c+d x)}{d}-\frac {a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac {\left (a^2+b^2\right ) \tan (c+d x)}{d} \]

[Out]

-3*a*b*arctanh(cos(d*x+c))/d-(2*a^2+b^2)*cot(d*x+c)/d-1/3*a^2*cot(d*x+c)^3/d+3*a*b*sec(d*x+c)/d-a*b*csc(d*x+c)
^2*sec(d*x+c)/d+(a^2+b^2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2990, 2702, 294, 327, 213, 3279, 459} \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) \tan (c+d x)}{d}-\frac {\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {3 a b \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a b \sec (c+d x)}{d}-\frac {a b \csc ^2(c+d x) \sec (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(-3*a*b*ArcTanh[Cos[c + d*x]])/d - ((2*a^2 + b^2)*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) + (3*a*b*Sec[c
+ d*x])/d - (a*b*Csc[c + d*x]^2*Sec[c + d*x])/d + ((a^2 + b^2)*Tan[c + d*x])/d

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2990

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[2*a*(b/d), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 3279

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.),
x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff^(n + 1)/f, Subst[Int[x^n*((a + (a + b)*ff^2*x^2
)^p/(1 + ff^2*x^2)^((m + n)/2 + p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/
2] && IntegerQ[n/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = (2 a b) \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+\int \csc ^4(c+d x) \sec ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right ) \left (a^2+\left (a^2+b^2\right ) x^2\right )}{x^4} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(2 a b) \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac {\text {Subst}\left (\int \left (a^2 \left (1+\frac {b^2}{a^2}\right )+\frac {a^2}{x^4}+\frac {2 a^2+b^2}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {(3 a b) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {3 a b \sec (c+d x)}{d}-\frac {a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac {\left (a^2+b^2\right ) \tan (c+d x)}{d}+\frac {(3 a b) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {3 a b \text {arctanh}(\cos (c+d x))}{d}-\frac {\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {3 a b \sec (c+d x)}{d}-\frac {a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac {\left (a^2+b^2\right ) \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.88 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\csc ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-4 \left (4 a^2+3 b^2\right ) \cos (2 (c+d x))+\left (8 a^2+6 b^2\right ) \cos (4 (c+d x))+3 b \left (2 b+10 a \sin (c+d x)-6 a \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (2 (c+d x))-6 a \sin (3 (c+d x))+3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))-3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))\right )\right )}{192 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]^3*(-4*(4*a^2 + 3*b^2)*Cos[2*(c + d*x)] + (8*a^2 + 6*b^2)*Cos[4*(c + d*x)]
+ 3*b*(2*b + 10*a*Sin[c + d*x] - 6*a*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[2*(c + d*x)] - 6*a*S
in[3*(c + d*x)] + 3*a*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] - 3*a*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)])))/(
192*d*(-1 + Cot[(c + d*x)/2]^2))

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+2 a b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\)	\(136\)
default	\(\frac {a^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+2 a b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\)	\(136\)
risch	\(\frac {6 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-4 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-10 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+\frac {32 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{3}+8 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+10 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-\frac {16 i a^{2}}{3}-4 i b^{2}-6 a b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(181\)
parallelrisch	\(\frac {72 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}+6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b +4 \left (5 a^{2}+3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (-5 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{2} \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a b \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (5 a^{2}+3 b^{2}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-108 a b}{24 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(181\)
norman	\(\frac {\frac {a^{2}}{24 d}+\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {5 \left (7 a^{2}+6 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (11 a^{2}+6 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (11 a^{2}+6 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (49 a^{2}+48 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {\left (49 a^{2}+48 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {7 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {19 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {33 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {3 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(312\)

[In]

int(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-1/3/sin(d*x+c)^3/cos(d*x+c)+4/3/sin(d*x+c)/cos(d*x+c)-8/3*cot(d*x+c))+2*a*b*(-1/2/sin(d*x+c)^2/cos(
d*x+c)+3/2/cos(d*x+c)+3/2*ln(csc(d*x+c)-cot(d*x+c)))+b^2*(1/sin(d*x+c)/cos(d*x+c)-2*cot(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 6 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, a^{2} + 6 \, b^{2} - 6 \, {\left (3 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(4*(4*a^2 + 3*b^2)*cos(d*x + c)^4 - 6*(4*a^2 + 3*b^2)*cos(d*x + c)^2 + 9*(a*b*cos(d*x + c)^3 - a*b*cos(d*
x + c))*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*(a*b*cos(d*x + c)^3 - a*b*cos(d*x + c))*log(-1/2*cos(d*x
+ c) + 1/2)*sin(d*x + c) + 6*a^2 + 6*b^2 - 6*(3*a*b*cos(d*x + c)^2 - 2*a*b)*sin(d*x + c))/((d*cos(d*x + c)^3 -
d*cos(d*x + c))*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**4*sec(d*x+c)**2*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, a b {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, b^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} - 2 \, a^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )}}{6 \, d} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(3*a*b*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x
+ c) - 1)) - 6*b^2*(1/tan(d*x + c) - tan(d*x + c)) - 2*a^2*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x
+ c)))/d

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.96 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {48 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {132 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a*b*tan(1/2*d*x + 1/2*c)^2 + 72*a*b*log(abs(tan(1/2*d*x + 1/2*c))) + 21*a
^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c) - 48*(a^2*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c
) + 2*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (132*a*b*tan(1/2*d*x + 1/2*c)^3 + 21*a^2*tan(1/2*d*x + 1/2*c)^2 + 12
*b^2*tan(1/2*d*x + 1/2*c)^2 + 6*a*b*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 11.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.87 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {20\,a^2}{3}+4\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (23\,a^2+20\,b^2\right )+\frac {a^2}{3}-34\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,a^2}{8}+\frac {b^2}{2}\right )}{d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {3\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int((a + b*sin(c + d*x))^2/(cos(c + d*x)^2*sin(c + d*x)^4),x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (tan(c/2 + (d*x)/2)^2*((20*a^2)/3 + 4*b^2) - tan(c/2 + (d*x)/2)^4*(23*a^2
+ 20*b^2) + a^2/3 - 34*a*b*tan(c/2 + (d*x)/2)^3 + 2*a*b*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 - 8*tan
(c/2 + (d*x)/2)^5)) + (tan(c/2 + (d*x)/2)*((7*a^2)/8 + b^2/2))/d + (a*b*tan(c/2 + (d*x)/2)^2)/(4*d) + (3*a*b*l
og(tan(c/2 + (d*x)/2)))/d